Question

If $A=\left[\begin{array}{ccc}i& 0& 0\\ 0& i& 0\\ 0& 0& i\end{array}\right]$ then $n\in N$, $A^{4n+1}=$

Answer 1

$A=\left[\begin{array}{ccc}i& 0& 0\\ 0& i& 0\\ 0& 0& i\end{array}\right]$

$=i\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$=iI$ where $I$ is a $3\times 3$ identity matrix

$\Rightarrow A^2=i^2{\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}^2$

$\Rightarrow A^2=-1{\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}^2=-{\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]}^2=-I$

$\Rightarrow A^3=A.A^2=iI.-I=-i{I}^2=-iI$

$\Rightarrow A^4=A^3.A=-iI,iI=-i^2{I}^2=I$

$\Rightarrow A^{4n}=I^n=I$

$\Rightarrow A^{4n+1}=I.A^1=IA=IiI=i{I}^2=iI=A$ is proved